Problem: Captain Nadia has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Tiffany and her merciless band of thieves. If her ship hasn't already been hit, Captain Nadia has probability $\dfrac{1}{2}$ of hitting the pirate ship. If her ship has been hit, Captain Nadia will always miss. If her ship hasn't already been hit, dread pirate Tiffany has probability $\dfrac{2}{7}$ of hitting the Captain's ship. If her ship has been hit, dread pirate Tiffany will always miss. If the Captain and the pirate each shoot once, and the Captain shoots first, what is the probability that the Captain misses the pirate ship, but the pirate hits?
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is the Captain missing the pirate ship and event B is the pirate hitting the Captain's ship. The Captain fires first, so her ship can't be sunk before she fires her cannons. So, the probability of the Captain missing the pirate ship is $\dfrac{1}{2}$. If the Captain missed the pirate ship, the pirate has a normal chance to fire back. So, the probability of the pirate hitting the Captain's ship given the Captain missing the pirate ship is $\dfrac{2}{7}$. The probability that the Captain misses the pirate ship, but the pirate hits is then the probability of the Captain missing the pirate ship times the probability of the pirate hitting the Captain's ship given the Captain missing the pirate ship. This is $\dfrac{1}{2} \cdot \dfrac{2}{7} = \dfrac{1}{7}$